(M-15) Raising one power to another

and a graphic use of logarithms

When (VP) itself is raised to the Qth power

What is [(102)]3 ? Obviously, we need multiply 10 by itself many times. Since

(102) = 10.10 = 100

[(102)]3 should multiply the above by itself for a total of 3 times:

(10.10).(10.10).(10.10) = 106

in all 2.3=6 factors of 10. In other words, if 10 is raised to the second power, and the result is then raised to the 3rd power, that is the same as raising 10 to the (2.3) = 6th power.

Indeed, the same argument can be made for any numbers, as long as they are raised to powers which are whole numbers. For any choice of whole numbers P and Q, if you raise some number V to power P and then raise the result to power Q, the result is like raising the number V to power (P.Q):

(VP)Q = V(P.Q)

For the next step we must decide--can one also raise V to a power that is not a whole number, and what does that mean? We proceed by stages: first show formal steps of how it is done, preserving the same rules known to exist for powers that are whole numbers. After that me may interpret the meaning of what was found.

Say V = 10P so that P = log V. Raising V to power Q and keeping the same rules found for integers P and Q

VQ = [10P]Q = 10(P.Q)

Taking the logarithm of both sides

log [VQ] = Q.P = Q log V

The logarithm of a number raised to the Qth degree is Q times its logarithm.

In this manner, given a number V, logarithms help calculate VQ, even if Q is not a whole number (as discussed below). The prescription:

---Take the numberV
---Find its logarithm P,
---Multiply P by Q to get Q log V. Say that is the number U.
--Find the number whose logarithm U is, that is, find 10U . That is the value of VQ .

But how does one interpret, raising a number to the power of an arbitrary real number?

(1) We may start with a simple example. Suppose

V.V = 10

Then V is the square root of 10 or √10, approximately 3,16227766...
If any powers a and b satisfy

(xa).(xb) = x(a+b)

then V behaves the way one expects 101/2 or 100.5 to do. Namely, multiplying if by itself:

(100.5).(100.5) = (101) = 10.

(2) Similarly for the Qth root of 10--the number which must be multiplied by itself Q times to get 10. If written as 101/Q, it would satisfy the equation for x(a+b), with 101/Q multiplied by itself Q times. (We do not go now into the question of how the Qth root is derived: there exist methods.)

(101/Q).(101/Q).(101/Q)... (Q times)... = (101) = 10

This can also be expressed using the earlier equation (in the box above). If 101/Q is raised to the Qth power and the earlier relation

(VP)Q = V(P.Q))

still holds, then

[101/Q]Q = 10 (Q /Q) = 101 = 1.

Furthermore, if the general relation holds for any two factors, let

V = 101/Q (so that log V = 1/Q)

Raising this to the Pth power

VP = [101/Q]P = 10(P/Q)

This lets one formally define the power (P/Q), i.e. any rational number: first raise V to power 1/Q (that is, take the Qth root of V), then raise all that to the Pth power, by multiplying it by itself P times.

(3) There is no good way to raise 10 to a power which is "irrational," a number which cannot be written as a fraction P/Q; for instance, raising 10 to power √2 or π . However, even though such numbers can never be exactly expressed as a fraction, there exist ways of approaching them through a series of fractions F1, F2, F3 ... which approximate them ever more closely; log π, for instance, is near 1022/7 and even closer to 10355/113, and by continuing this sequence (or more conventionally, using a sequence of decimal fractions), one can approximate it as closely as one wishes.

In general, the numbers N1, N2, N3...equal to 10 raised to the powers F1, F2, F3 approaching the irrational number will also get closer to each other. One guesses that if the process is taken far enough, the result represents 10 to the irrational power as accurately as one may wish.
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We already know of course that Kepler found T2 was proportional to R3 (so columns 4 and 5 should be equal, except the values used here are not accurate), Kepler's famous 3rd law. It can be written

T = K R(3/2)

with K some number that depends on the units used for T and R. But suppose we are in Kepler place--all we have are the tables on the left, giving corresponding entries of T and R (ignore now the fact that Kepler knew nothing of the last 3 planets listed!). How can one find the relationship? Plotting T against R just gives some curve: is there a way of finding what curve it is? Logarithms provide such a way. Suppose we guess the curve satisfies a relationship of the form

T = K RN

Taking logarithms of both sides (if the sides are equal, their logarithms are equal too)

log T = log K + log (RN)

Or

log T = log K + N log R

Now plot an (x, y) system of coordinates, and for each planet, assign x = log R and y=logT
The equation of the line y against x is then

y = log K + N x

If our guess of the formula T = K RN was accurate, we get a straight line y = b + Nx and the slope of the line is N; that is, given any two points on it, (x1, y1) and (x2, y2), then

N = (y1 – (y2) /
(x1 – (x2)

If the guess was wrong, the line will not be straight. Note that the choice of units for T (years, seconds...) and R (astronomical units, kilometers, miles...) does not matter. If the units used for measuring T change, T is replaced by qT, where q is some constant number. Then log T is replaced by (log q+logT) and therefore just modifies the value of b = log K. In the table above, using the period of Earth and its mean distance as units, the point contributed by Earth is (x,y) = (0,0) ; therefore b=0, and taking the origin as the second point simplifies the slope equation to

N = y / x

where (x, y) is any point on the graph. You are encouraged to draw the graph and convince yourself that the exponent is indeed 3/2.

Exploring Further

Other fractional exponents arise from the gas laws. You may have learned that in a gas the pressure P of a given quantity of gas (say, one gram) is inversely proportional to its volume V

P V = constant

When the gas is compressed to half its volume, its pressure doubles.

That, however, only holds true if the temperature stays the same. In fact, when you pump gas into a container of half the volume, it not only generates higher pressure, but you invest energy overcoming the pressure of the gas already there. As a result (as users of bicycle pumps know well) the gas heats up, and its pressure increases more than twice. It turns out that if heat is not allowed to flow in or out, a good approximation for the gas law is

P Vγ = constant

Where γ (the Greek lower-case g) is about 7/5 for air and for gases where two identical molecules combine (nitrogen, oxygen, hydrogen and chlorine), and about 5/3 for "monoatomic" gases such as helium, whose atoms stay solitary. Such "adiabatic gas laws" are especially important in big systems (e.g. large-scale motions in the atmosphere) because as the volume increases there is less and less surface area through which each unit of volume can gain or lose heat, making it harder for much heat to flow in or out.